# Calculate the effective access time

This question seems to be causing a lot of debate and I'm wondering whether my working is correct.

A computer with a single cache (access time 20ns) and main memory (access time 500ns) also uses the hard disk (average access time 0.01ms) for virtual memory using paging. If it is found that the cache hit rate is 95% and the page fault rate is 1%
Calculate the effective (average) access time (EAT) of this system for a sequential access system.

(a) 73.5ns. (b) 55ns. (c) 50ns. (d) 74.75ns.

My working was as follows: (0.95x20) + 0.05 (0.01(10000+500+20) + 0.99(500+20))

Leading me to an answer of 50 being C), however others have seem to have chosen D) as their answer

Any help would be appreciated.

• hard disk (average access time 0.01ms) WHOA! (Or solid state, but the cache access time doesn't look the type.) Jan 10, 2021 at 17:42